3

why does the Wegscheider condition or detailed balance have to be true for chemical reactions and what does it mean to violate it in terms of thermodynamics?

here is an attempt to derive a simple example that violates the condition. there are 3 chemical species A, B, C that follow these reversible reactions and obey mass action kinetics:

$A \rightarrow^{k_1} B$

$B \rightarrow^{k_2} A$

$B \rightarrow^{k_3} C$

$C \rightarrow^{k_4} B$

$C \rightarrow^{k_5} A$

$A \rightarrow^{k_6} C$

Detailed balance says that the products of rate constants along a cycle must equal 1. one can draw a cycle here: $A$ goes to $B$ which goes to $C$ (this is one part of the cycle) and then $C$ goes back to $A$ (the other part of the cycle). detailed balance requires that:

$\frac{k_1}{k_2}\frac{k_2}{k_3}\frac{k_5}{k_6} = 1$

suppose all the forward rates are 2 and the reverse rates 1: $k_1 = k_3 = k_5 = 2, k_2 = k_4 = k_6 = 1$. then:

$\frac{k_1}{k_2}\frac{k_2}{k_3}\frac{k_5}{k_6} = \frac{2}{1}\frac{2}{1}\frac{2}{1} = 6 \neq 1$

(rate constants that satisfy detailed balance would be, eg: $k_1 = 2, k_2 = 1, k_3 = 2, k_4 = 1, k_5 = 4, k_6 = 1$)

questions:

  1. why is this thermodynamically inconsistent?
  2. what does it mean for the existence of equilibrium?

it looks like there is still equilibrium when the rates are equal, ie when:

$[A]k_1 = [B]k_2$

$[B]k_3 = [C]k_4$

$[C]k_5 = [A]k_6$

where is the problem in having equilibrium here?

Qmechanic
  • 220,844
bal
  • 31

1 Answers1

2

For your cycle, if you break detailed balance you will have currents along the cycle, and the presence of currents in your system will be inconsistent with equilibrium assumption (no flow of energy or particles in the system).

That did not prevent the system to reach a steady state, but this will be a non-equilibrium steady state.

Note that there is a strong difference between equilibrium and steady state (presence of currents), even if both case feature an absence of evolution of the concentration.

Edit:

Currents correspond to a flow of molecules due to non-equal forward and backward rates. If you try to solve the equilibrium equation, the first one will give you $$ [B] = \frac{k_1}{k_2}[A] $$ you can insert it into the second one to get $$ [C] = \frac{k_3 k_1}{k_4 k_2}[A] $$ and finally $$ [A] = \frac{k_5 k_3 k_1}{k_6 k_4 k_2} [A]. $$ Therefore you have a possible equilibrium solution (except for zero concentration solution) only when you have detailed balance.

However you still have non-equilibrium steady state solution by solving $$ \frac{\mathrm d }{\mathrm d t}[A] = k_2 [B] - k_1[A] +k_5 [C]-k_6[A] = 0$$

Hadrien
  • 476